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Inverse Trigonometric Functions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsInverse Trigonometric Functions2 Marks
Question
Evaluate the following:
$\sec^{-1}\Big(\sec\frac{9\pi}{5}\Big)$
✓

Answer

We have
$\sec^{-1}\Big(\sec\frac{9\pi}{5}\Big)=\sec^{-1}\Big[\sec\Big(2\pi-\frac{\pi}{5}\Big)\Big]$
$=\sec^{-1}\Big[\sec\Big(\frac{\pi}{5}\Big)\Big]$
$=\frac{\pi}{5}$

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